Chemistry - Ionic Equilibrium Question with Solution | TestHub

Weak monoacidic base, e.g. $\mathrm{BOH}$ is neutralised. $$\mathrm{BOH}+\mathrm{HCl}⟶\mathrm{BCl}+{\mathrm{H}}_2\mathrm{O}$$ At equivalence point all $\mathrm{BOH}$ gets converted into salt and remember! the concentration of ${\mathrm{H}}^{+}$(or $\mathrm{pH}$ of solution) is due to hydrolysis of resultant salt ( $\mathrm{BCl}$, cationic hydrolysis here) $$\underset{\mathrm{C}(1-h)}{{\mathrm{B}}^{+}}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \underset{\mathrm{Ch}}{\rightleftharpoons }\mathrm{OH}+\underset{\mathrm{Ch}}{{\mathrm{H}}^{+}}$$ Volume of $\mathrm{HCl}$ used up, $$V_a=\frac{N_b{V}_b}{N_a}=\frac{2.5\times 2\times 15}{2\times 5}=7.5\mathrm{mL}$$ Concentration of salt, $$\begin{array}{rl}[\mathrm{BCl}]& =\frac{\text{ Concentration of base }}{\text{ Total volume }}=\frac{2\times 25}{5(7.5+2.5)}=\frac{1}{10}=0.1\\ K_h& =\frac{C{h}^2}{1-h}=\frac{K_w}{K_b}\end{array}$$ ( $h$ should be estimated whether that can be neglected or not) on calculating $h=0.27$ (significant, not negligible) $$\left[{\mathrm{H}}^{+}\right]=Ch=0.1\times 0.27=2.7\times 10^{-2}\mathrm{M}$$ When $\left[{\mathrm{H}}^{+}\right]$is asked to calculate in connection with neutralisation, it Should be calculated: Before neutralisation Using Oslwald's dilution law During neutralisation Considering buffer solution Half neutralisation $\mathrm{pH}=\mathrm{p}{K}_a$ and $\mathrm{p}{K}_b$ At the end of neutralisation Considering hydrolysis of salt